Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct minimumi {
  basic_string<int> a;
  minimumi operator+(const minimumi& b) const {
    basic_string<int> tmp(a.size() + b.a.size(), 0);
    merge(a.begin(), a.end(), b.a.begin(), b.a.end(), tmp.begin());
    tmp.resize(min((int)tmp.size(), 10));
    return {tmp};
  }
};
minimumi v[100005][18];
int p[100005][18], d[100005], n;
basic_string<int> e[100005];
void dfs(int x, int par) {
  p[x][0] = par;
  for (int i = 1; i < 18; i++) {
    int y = p[x][i - 1];
    p[x][i] = p[y][i - 1];
    v[x][i] = v[x][i - 1] + v[y][i - 1];
  }
  for (int y : e[x]) {
    if (y == par) continue;
    d[y] = d[x] + 1;
    dfs(y, x);
  }
}
int dizi(int x, int k) {
  for (int i = 0; i < 18; i++)
    if (k & (1 << i)) x = p[x][i];
  return x;
}
minimumi racunaj(int x, int k) {
  minimumi w;
  for (int i = 0; i < 18; i++)
    if (k & (1 << i)) {
      w = w + v[x][i];
      x = p[x][i];
    }
  return w;
}
int lca(int x, int y) {
  if (d[x] > d[y]) swap(x, y);
  y = dizi(y, d[y] - d[x]);
  if (x == y) return x;
  for (int i = 17; i >= 0; i--) {
    int xx = p[x][i];
    int yy = p[y][i];
    if (xx != yy) {
      x = xx;
      y = yy;
    }
  }
  return p[x][0];
}
void ans(int x, int y, int a) {
  int z = lca(x, y);
  auto w = racunaj(x, d[x] - d[z]) + racunaj(y, d[y] - d[z]) + v[z][0];
  a = min(a, (int)w.a.size());
  w.a.resize(a);
  cout << a;
  for (int x : w.a) cout << ' ' << x;
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cerr.tie(nullptr);
  int m, q;
  cin >> n >> m >> q;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    e[u] += v;
    e[v] += u;
  }
  for (int i = 1; i <= m; i++) {
    int x;
    cin >> x;
    v[x][0] = v[x][0] + minimumi{{i}};
  }
  dfs(1, 1);
  while (q--) {
    int x, y, a;
    cin >> x >> y >> a;
    ans(x, y, a);
  }
}